Slide 1

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 2

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 3

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 4

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 5

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 6

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 7

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 8

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 9

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 10

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 11

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 12

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 13

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 14

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 15

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 16

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 17

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 18

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 19

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 20

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 21

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 22

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 23

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 24

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 25

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 26

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 27

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 28

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 29

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 30

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 31

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 32

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 33

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 34

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 35

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 36

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 37

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 38

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 39

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 40

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 41

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 42

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 43

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 44

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 45

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 46

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 47

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 48

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 49

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 50

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 51

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 52

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 53

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 54

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 55

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 56

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 57

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 58

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 59

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 60

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 61

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 62

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 63

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 64

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 65

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 66

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 67

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 68

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 69

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 70

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 71

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 72

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 73

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 74

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 75

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 76

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 77

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 78

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 79

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 80

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 81

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 82

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 83

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 84

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 85

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 86

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 87

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 88

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 89

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 90

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 91

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 92

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 93

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 94

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 95

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 96

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 97

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 98

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 99

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 100

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 101

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 102

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 103

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 104

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 105

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 106

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 107

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 108

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 109

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 110

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 111

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 112

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 113

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 114

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 115

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 116

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 117

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 118

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 119

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 120

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 121

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 122

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 123

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 124

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 125

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 126

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 127

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 128

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 129

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 130

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 131

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 132

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 133

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 134

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 135

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 136

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 137

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 138

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 139

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 140

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 141

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 142

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 143

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 144

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 145

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 146

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 147

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 148

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 149

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 150

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 151

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 152

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 153

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 154

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 155

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 156

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass


Slide 157

Equations:
Linear and
Systems

Overview
In this module you will be learning 3 topics all involving linear
equations. Linear Equations are equations with one or more
variables whose power is of degree one. If the degree of any of
the variables is not of degree one, then it is not a linear equation.

Three Topics:
Solving Linear Equations for a given variable
Finding and using the Equation of a Line
Solving and using Systems of Linear Equations
*You will be able to navigate easily to the topics of interest by clicking
on the links on each page.
*The icon of a house in the upper right-hand corner of each page will
return you to this page.
*If you click on the title of any page, that will return you to that
particular sub-topic of the module.

Solving Linear Equations
•
•
•
•
•
•
•

Writing Equations Using Application Problems
Solving Equations by Using Addition and Subtraction
Solving Equations by Using Multiplication and Division
Solving Equations Using Algebra Tiles
Solving Multi-Step Equations
Solving Equations with the Variable on Each Side
Application Problems

Writing Equations
• First, we need to be able to translate our words in
mathematical problems.
• Here are basic terms which determine which
mathematical operations to use.
Addition

Subtraction

Multiplication

Division

Total

Minus

Times

Divided by

Plus

Less than

Of

Per

Added to

Subtracted from

Each

Ratio

More than

Decreased by

Factors

Rate

Sum(use
parentheses to
represent the
answer.)

Difference (use
parentheses to
represent the
answer.)

Product (use
parentheses to
represent the
answer.)

Quotient (use
parentheses to
represent the
answer.)

Writing Equations - Example
• You want to purchase an iPod and you know that it costs
about $160. You have $60 dollars saved all ready. Your dad
tells you that he will pay you $20 each week for mowing the
grass. How long will it take you to earn the additional money?
• 1st- You know that you have $60 all ready. As a result, you
need to earn $160-$60 = $100.

Writing Equations-continued
2nd- You will earn $20 each week and you need to earn $100.
Let x = the number of weeks you need to mow the lawn
$20 x = 100
Let’s find what number you have to multiply 20 by to get to 100?
X = 5
It will take you 5 weeks to earn the $100 which are left to purchase
the iPod.
Lets check your answer: $160 - $60 = $100 left to earn
 $20 ( 5) = $100
$100 + $60 = $160.

Writing Equations-continued
• Click on the link below to practice writing equations.
The beginning of the link will provide you with examples. If you
continue to scroll down, you will be able to work with 5
problems by clicking on your answer choice and the computer
will give you immediate feedback as to whether your choice is
correct or not.
http://www.mathgoodies.com/lessons/vol7/equations.html

Solving Equations by
Addition and Subtraction
• When we balance equations, we have to remember to share
the same amount with both sides. This is done by using the
Addition and Subtraction Properties of Equality.
• x + 11= 15
• x+11 -11 = 15 – 11 Subtract 11 from both sides so that the
variable is the only term on the left side of
the equal sign
x+0=4
“x” plus 0 is equal to “ x”
x=4
Simplify

Solving Equations by Addition
and Subtraction-continued
• To practice solving equations using addition and subtraction
problems, open the below link.
• Once you click on the link, the first thing you will see is an
example. If you scroll down you will see two choices; one
choice is to review more examples and the second choice is
for you to test yourself by working similar problems.

• http://www.sosmath.com/algebra/solve/solve1/s11/s11.html

Solving Equations by Using
Multiplication and Division
• Review: 5x=20 therefore x has to be 4 because we know
that 5(4) = 20. If we did not know that 5(4)=20, using 20
and 5 which operation would we have to perform to get
4? We would have to divide 20 by 5. As a result, division
will undo multiplication and multiplication will undo
division. These are called the Multlipication and the
Division Properties of Equalities. Let’s apply this concept
to solving equations.

Solving Equations by Using
Multiplication and Division-

continued

• Solve:
• -13 x = 52
• -13 x = 52 Divide both sides by -13 because
-13
-13 division will undo multiplication
• X= -4
• Let’s check to see if this is correct.
• -13(-4) = 52 therefore x = -4.

Solving Equations by Using
Multiplication and Division-

continued

•
•
•
•

Solve
-5h = -2/3
-5h = -2/3 Divide both sides by -5
h= -2/3 (-1/5) Divide fractions by multiplying
by its reciprocal
• h=2/15
Check: -5(2/15) =-10/15= -2/3
Let’s solve -5h = -2/3 by multiplying each side of this equation
by the reciprocal of h’s coefficient. The reciprocal of -5 is -1/5
therefore:
-1/5 (-5h) = -1/5 (-2/3) Recall the product of a number and it’s
h= 2/15
reciprocal is equal to 1.

Solving Equations by Using
Multiplication and Division-

continued

• To practice solving equations using multiplication and division
problems, open the below link.
• There are four steps once you click on the link. The first step
will show you how to balance the equation. Step 2 will give
you an in-depth explanation of how to solve equations. Step 3
will show you 5 additional examples. Step 4 will allow you to
practice solving equations.
• http://www.math.com/school/subject2/lessons/S2U3L3GL.ht
ml

Solving Equations Using
Algebra Tiles
This link includes a video tutorial of how to use algebra tiles to
solve equations.
• http://www.youtube.com/watch?v=CpnzNmw1Mg8

Solving Multi-Step Equations
• There are several steps you will need to follow to solve multi-step
equations. Think of using these steps as you would use the order of
operations.
• Step 1: Use the Distributive Property to remove the grouping symbols.
• Step 2: Simplify the expressions on each side of the equal sign by
combining like terms.
• Step 3: Combine like terms on different sides of the equal sign to get all
the variables on the same side and all the numbers without variables on
the other side together. Use what you learned in the first section for
solving addition and subtraction equations. ( The addition and/or the
subtraction properties of equality.)
• Step 4: Simplify each expression on each side of the equal sign.
• Step 5: Use what you learned in the second section for solving
multiplication and division equalities. (The multiplication and /or the
division property of equality.)

Solving Equations with the
Variable on Each Side
• The key is to combine like terms by using the previous
strategies you learned from solving equations.
• Solve for f.
-16f – 2 = -15f + 17
-f – 2 = 17 Add 15f to both sides
-f = 19
Add 2 to both sides
f = -19
Multiply both sides by -1

Solving Equations with the
Variable on Each Side-continued
• Open the below link to practice solving multi-step equations
and equations with variables on both sides of the equal sign.
• Once you open the link you will be able to practice solving 15
multi-step equations with there solutions at the bottom of the
website.
• http://www.education.com/study-help/article/solvingmultistep-algebraic-equations_answer/

Solving Multi-Step Equations
Using Algebra Tiles
• This link includes a video tutorial of how to use algebra tiles to
solve multi-step equations.
• http://www.youtube.com/watch?v=l00CeulzdZo
• To practice solving equations using algebra tiles click on the
below link. The link allows you to represent the equations by
clicking on the algebra tiles and move them around to solve
the equations.
• http://media.mivu.org/mvu_pd/a4a/homework/applets_two_
step.html

Finding the Equation of a Line
• This unit will cover basic graphing of
points and lines along with finding the
slope of lines given two points and from
the line graph. It will also cover how to
find the equation of lines from the lines
graph and given conditions. The unit
concludes with finding the equations of
parallel and perpendicular lines.

Finding the equation of a line
•
•
•
•
•
•
•
•
•
•
•
•

Review Graphing points
Slope from a given line
Slope between two points(Slope Formula)
Graphing lines with Tables, Point & Slope and Slope Intercept
Finding equation given slope and y-intercept
Finding equation given point and slope(Slope Intercept and
Point Slope)
Finding equation of the line given graph
Finding equation given two points
Standard form ( Intercepts and putting into Slope Intercept)
Parallel Lines
Perpendicular Lines
Applications

Review Graphing Points
Points are (x,y) ordered pairs that give a
position on a coordinate system.

y-axis
Quadrant II

Quadrant I

x-axis

Quadrant III

Quadrant IV Back to section
Title page

To graph a point on a coordinate system
You must travel left or right on the x-axis
and the up or down on the y-axis.

Right
Left

Up
Down
Back to section
Title page

The Finding the equation of a line

(0,0) is called the origin.

Back to section
Title page

Points in Quadrant I have positive x
values and positive y values
y

Quadrant I

(3,4)
Right 3
Up 4

x

Back to section
Title page

Points in Quadrant II have negative
x values and positive y values
Quadrant II

(-5,6)
Left 5
Up 6

Back to section
Title page

Points in Quadrant III have negative
x values and negative y values

(-7,-3)
Left 7
Down 3
Quadrant III
Back to section
Title page

Points in Quadrant IV have positive
x values and negative y values

(8,-5)
Right 8
Down 5
Quadrant IV
Back to section
Title page

Practice Graphing Lines
• Go to this website if you would like to practice graphing points
• http://www.webmath.com/gpoints.html

Back to section
Title page

Slope of a given line
•In this section you will review
how to find the slope of a given
line from the graph of the line.

Back to section
Title page

Slope(m) is defined as the vertical
change of the line divided by the
horizontal change of the line.

Rise is the vertical change
Run is the horizontal change

𝑚=

𝑅𝑖𝑠𝑒
𝑅𝑢𝑛
Back to section
Title page

Slope of a given line
Identify two points on the given line

(1,1) and (5,4)
𝑅𝑖𝑠𝑒
𝑅𝑢𝑛

𝑚=
m=

3

3
5

5
Back to section
Title page

Slope of a given line

(-4,7) and (5,2)
-5
9

-5
m=
9
Back to section
Title page

Slope of a given line

(-6,1) and (4,5)
5
m=

5

10
1
m=
2

10
(Simplify)

Back to section
Title page

Slope of a given line

No rise

m=

0

11

11
m=0

The black line above is called a horizontal line.
Horizontal lines always have a slope of 0.
Back to section
Title page

Slope of a given line

3
m=
0
m = undefined

3
No Run

The black line above is a vertical line. A
vertical line has an undefined slope.
Back to section
Title page

Click on the link below to
practice finding the slope of a line
http://www.mathopenref.com/coordslope.html

Back to section
Title page

Finding Slope of Line
(Slope Formula)
• In this section you will review how to
use the Slope Formula to find the
slope of a line between two points.

Back to section
Title page

Given two points
1st Point

2nd Point

Slope Formula
Back to section
Title page

Find the Slope between
(3,4) and (5,7)

7 − 4
𝑚=
5 − 3
3
𝑚=
2

Back to section
Title page

Find the Slope between
(6,-5) and (-2,7)
7 − (−5)
𝑚=
−2 − 6

12
m=
−8
3
m=
−2

Back to section
Title page

Find the Slope between
(5,4) and (-1,10)

10 − 4
𝑚=
−1 − 5
6
m=
−6
m= −1

Back to section
Title page

Find the Slope between
(8,4) and (-3,4)
4 − 4
𝑚=
−3 − 8
0
𝑚=
−11
𝑚=0
This is a horizontal line

Back to section
Title page

Find the Slope between
(-2,6) and (-2,9)
9 −6
𝑚=
−2 − (−2)
3
𝑚=
0
𝑚 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑
This is a vertical line.

Back to section
Title page

Go to this website if you would
like to do a worksheet on finding
The slope between two points.
http://www.kutasoftware.com/Free
Worksheets/Alg1Worksheets/Slope
%20From%20Two%20Points.pdf

Graphing lines
• In this section you will review how
to graph lines.
• Using Tables
• Using a Point and the Slope
• Using the Slope and the y intercept
Back to section
Title page

Using Tables
Plot the points and connect them.
x
-3
-2
-1
0
1
2
3
.

y
-10
-8
-6
-4
-2
0
2

Every point
on the line is a
solution to the
same line
equation

You need only two
points to graph the line
Back to section
Title page

Using a Point and the Slope
Plot the given Point and use the Slope to find the next point

Point
(1,2)

3 up
m=
4 right

or
3 down
m=
4 left

Back to section
Title page

Using Slope and y-intercept

y-intercept is the point where the
line goes through the y-axis

Back to section
Title page

Using Slope and y-intercept
y-intercept = 4
-3 down
m=
5 right
3 up
m=
-5 left
Back to section
Title page

Finding equation given slope and
y-intercept
•In this section you will
review how to find the
equation of a line given the
slope of the line and the yintercept.
Back to section
Title page

Slope Intercept Form
y = mx + b
where
m is the slope
and
b is the y-intercept
Back to section
Title page

Find the equation of the line
Given a slope of 5 and y-intercept of 6

m=5

b=6

y = mx + b
(substitute for m and b)

y = 5x + 6
Back to section
Title page

Find the equation of the line
Given a slope of 8 and y-intercept of -3

m=8

b = -3

y = mx + b
(substitute for m and b)

y = 8x - 3
Back to section
Title page

Find the equation of the line

Given:𝑚 =

3
− 𝑎𝑛𝑑
4

𝑏=6

y = mx + b
(substitute for m and b)

3
𝑦 =− 𝑥+6
4
Back to section
Title page

Finding equation given
point and slope
•In this section you will review
how to find the equation of a
line given the slope and a
point.

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 6 and point (2,3)

y = mx + b
[substitute for m and (x,y)]

3 = 6(2) + b
3 = 12 + b
-9 = b
(substitute for m and b)

y = 6x - 9
Back to section
Title page

Find the equation of the line
(Slope Intercept)
2
𝐺𝑖𝑣𝑒𝑛: 𝑚 = 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (6,7)
3
y = mx + b
[substitute for m and (x,y)]
7=

2
3

6 +𝑏

7=4+𝑏
3=𝑏
(substitute for m and b)

y

2
= 𝑥
3

+3

Back to section
Title page

Find the equation of the line
(Slope Intercept)
3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = − 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡 (−4,2)
5

y = mx + b
[substitute for m and (x,y)]
3
2 =- −4 +
5
12
2= + 𝑏
5
2
= 𝑏
5

𝑏

(substitute for m and b)

y

3
=- 𝑥
5

+

2
5

Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = 0 and point (-2,-6)
y = mx + b
[substitute for m and (x,y)]

-6 = 0(-2) + b
-6 = b
(substitute for m and b)

y = 0x – 6
y=-6
(This is the equation of a horizontal line) Back to section
Title page

Find the equation of the line
(Slope Intercept)
Given: m = undefined and point (-2,-6)
Since the slope is undefined, you know that the
line is a vertical line. Vertical lines go through
the x-axis at the x value of the ordered pair.

The equation is

x = -2
Back to section
Title page

Point Slope Form
𝑦 − 𝑦1 = m(x - 𝑥1 )
You can use this form when you are
given a point and a slope of the line.
You will substitute for slope(m) and
point for 𝑥1 𝑎𝑛𝑑 𝑦1

Back to section
Title page

Point Slope Form
Find the equation of the line
Given: m = 7 and (2,4)

𝑦 − 𝑦1 = m(x - 𝑥1 )
y – 4 = 7(x – 2)
(This is an acceptable answer but you can also simplify it)

y – 4 = 7x - 14
y = 7x - 10
Back to section
Title page

Point Slope Form
Here is a website to visit for more
practice on Point Slope form
http://www.mathsisfun.com/algebra/lineequation-point-slope.html

Back to section
Title page

Finding equation of the line given graph

•In this section you will review
how to find the equation of the
line given the graph of the line

Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b = -3
Find the slope(m)
3
m=
7

7

3

(substitute for m and b)

y

3
= 𝑥
7

−3
Back to section
Title page

Finding equation of the line given graph
y = mx + b
Find the y-intercept (b)

b=4
Find the slope(m)
m=

-2
7

-2
7

(substitute for m and b)

y=

−2
𝑥
7

+4
Back to section
Title page

Finding equation of the line given graph
y-axis

This is a horizontal line.
It goes through the y-axis at 2.

The equation is
y=2

Back to section
Title page

Finding equation of the line given graph

This is a vertical line.
It goes through the x-axis at 5.

x-axis

The equation is
x=5
Back to section
Title page

Finding the equation of given
two points
•In this section you will review
how to find the equation of a
line when you are given two
points on the line.

Back to section
Title page

Finding the equation of given two points
Given: (2,3) and (5,8)
1st Find the slope between the two points
𝑚=

𝑦2 − 𝑦1
𝑥2 − 𝑥1

𝑚=

8 −3
5 −2
5
𝑚=
3

2nd Use the slope and one of the points
5
𝑚=
3

and (2,3)

y = mx + b
5
2 +𝑏
3
10
3=
+𝑏
3
3=

1
𝑏=−
3
(substitute for m and b)

5
1
𝑦= 𝑥 −
3
3

Back to section
Title page

Standard form of equations of lines
• In this section you will review the
standard form of the equation of a
line. You will also learn how to find
the x and y intercepts. You will also
change from slope intercept to
standard form and from standard form
to slope intercept.
Back to section
Title page

The standard form of the equation of
a line has both x and y on the same side
and the constant on the other with no
fractions present.

Ax + By = C
A, B, and C must be integers
The coefficient of x (A) must be positive
(and must be the first term in equation)
C is the constant term
Back to section
Title page

Finding the intercepts from standard form
3x + 4y = 12
To find the x intercept To find the y intercept
Let y = 0
Let x = 0
3x + 4(0) = 12
3(0) + 4y = 12
3x = 12
4y = 12
x=4
y=3
Plot the x and y
intercepts and
connect with the line.

Back to section
Title page

Changing from standard to slope intercept

3x + 4y = 16
-3x
-3x

Get y on a
side by
itself

4y = 16 – 3x
4

4

4

3
𝑦 =− 𝑥+4
4

Slope Intercept
form

Back to section
Title page

Changing from slope intercept to standard

2
𝑦 = 𝑥−8
3

Multiply each
term by 3 to
remove the
fraction

3 2
(3)𝑦 =
𝑥 − (3)8
1 3
3𝑦 = 2𝑥 − 24
-3y +24 -3y +24 2x – 3y = 24
24 = 2x – 3y

Back to section
Title page

Parallel Lines
•In this section you will review
how to find the equations of
parallel lines. You will review
how to find parallel lines to a
given line and when you are
given a slope.
Back to section
Title page

Parallel lines do not intersect.
Parallel lines have the same slope
Both lines have
a slope of

−7
𝑚=
9
The arrows on the lines indicate that the
lines are parallel

Back to section
Title page

Line parallel to given line
through the given point
You will now use
the given point
and the slope you
found to find the
next point

(5,8)

-3
4

-3
4

You must first
find the slope of
(red)given line

𝑟𝑖𝑠𝑒
𝑚=
=
𝑟𝑢𝑛

-3

Continue to
the next page
to find the
new line’s
equation

4
Back to section
Title page

Finding Parallel Line Equation
(5,8)

-3
4

-3

You must use the
given point(5,8)
the slope of
(red)given line

4

y = mx + b
𝑚=

−3
4

and point (5,8)

−3
5=
5 +𝑏
4

−15
5=
+𝑏
4
−3
35
= 𝑏 𝑎𝑛𝑑 𝑚 =
4
4
(Replace for m and b)

𝑦=

−3
𝑥
4

+

35
4

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line that is parallel to the
line y = 2x + 7 going through the point (3,4)
Parallel lines have the same slope so you must find the slope of the given line

Given m = 2 so the parallel m = 2
Parallel m = 2 Point is (3,4)
use y = mx + b
4 = 2(3) + b
4=6+b
-2 = b
(substitute for m and b)

So parallel line is y = 2x - 2

Back to section
Title page

Line parallel to given line
through the given point
−4
5

Parallel 𝑚 =
through (-7,8)
Use y = mx + b
−4
8=
−7 + 𝑏
5
28
8=
+𝑏
5
12
=𝑏
5
(substitute for m and b)

𝑦=

−4
𝑥
5

12
+
5

Back to section
Title page

Line parallel to given line
through the given point
Find the equation of the line parallel to 7x + 2y = 10
and going through point (3,-5)
1st Find the
slope from
given
equation

−7
𝑚=
2

point (3,-5)

y = mx + b
−7
3 +𝑏
2
−21
−5 =
+𝑏
2
11
=𝑏
2

−5 =
7x + 2y = 10
2y = 10 – 7x
7

𝑦 = 5 − 2x
𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑚 =

−7
2

(substitute for m and b)
−7
11
𝑦=
𝑥+
2
2
Back to section
Title page

All horizontal lines are parallel

y=6
y=3

y = -4

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

y=7

y = -2
Given
(0,-2)

Back to section
Title page

All vertical lines are parallel

x = -8

x = -1

x=3

Back to section
Title page

Find the equation of the line parallel to the
given line(red) through the given point(blue)

Given
(4,3)

x = -5

x=4

Back to section
Title page

Perpendicular Lines
• In this section you will review how
to find the equations of
perpendicular lines. You will
review how to find perpendicular
lines to a given line and when you
are given a slope.
Back to section
Title page

Perpendicular lines have slopes
that are negative reciprocals.
3
5
Red Line
5
𝑚=
3

Blue Line

-3

𝑚=−

5

3
5

Back to section
Title page

If you multiply the slopes of two
lines together and the result is -1,
then the two lines are perpendicular.
1st Line
𝑚=

2nd Line

5
3

5
3

𝑚=−

3
−
5

3
5

= -1

Back to section
Title page

Finding Perpendicular Line through given point
Black Line
5
𝑚=
8

Perpendicular Line
𝑚=−

8
5

Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

𝐺𝑖𝑣𝑒𝑛: 𝑦 = 4𝑥 + 8 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (2,5)
Slope of given line m = 4 so the perpendicular slope is

𝑚=

1
−
4

and point (2,5)

y = mx + b
1
2 +𝑏
4
1
5=− +𝑏
2

5=−

11
=𝑏
2

(substitute for m and b)
1
11
y=− 𝑥+
4
2
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

3
𝐺𝑖𝑣𝑒𝑛: 𝑦 = 𝑥 − 5 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (−2,6)
4
𝐺𝑖𝑣𝑒𝑛 𝑚=

3
4

𝑚=−

𝑠𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑚 = −
4
3

4
3

and point (-2,6)

y = mx + b
4
6 = − −2 + 𝑏
3
8
6= +𝑏
3
10
=𝑏
3

(substitute for m and b)
4
10
y=− 𝑥+
3
3
Back to section
Title page

Finding equation of line perpendicular to given line
through the given point

Given : 3x + 4y = 7 and going through (4,-6)
1st Find the
slope from
given
equation

Use 𝑚 =

4
3

𝑎𝑛𝑑 (4. −6)

y = mx + b
4
4 +𝑏
3
16
−6 =
+𝑏
3
16
−6 =
+𝑏
3
−34
= 𝑏
3
(substitute for m and b)
−6 =

4y = 7 – 3x
7
4

𝑦= -

3
𝑥
4

3
𝐺𝑖𝑣𝑒𝑛: 𝑚 = −
4
4
𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 ∶ 𝑚 =
3

4
34
𝑦= 𝑥 −
3
3

Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=6

(5,-3)

Perpendicular Line
x=5
Back to section
Title page

Find the equation of the line perpendicular to the
given line(red) through the given point(blue)

y=3

(-4,3)

x=5
Back to section
Title page

Applications

•In this section you will find
sample problems that are
applications of equations of
lines.

Back to section
Title page

Tennis Lessons

Click
here
for
answer

The graph above charts the cost of tennis lessons at a local tennis club. The
cost includes a one time membership fee and price per hour of the lessons
1. Find the linear equation that fits given chart.

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

2. How much will the total cost for 52 lessons be?

Back to section
Title page

Tennis Lessons-continued

Click
here
for
answer

3. If John has $850, how many lesson can he buy?

Back to section
Title page

Answers to Application Problem 1
1. Linear Equation that fits the chart. Two points that are on the line are
(0,150) and (15,300) so you can find the slope of the line.
Return to
300 −150 150
m = 15 −0 = 15 = 10
(This is the cost per lesson.)
problem
The y intercept(b) is 150. (This is the one time membership fee.)
The Linear Equation is y = 10x + 150

2. How much will the total cost for 52 lessons be?
Substitute 52 in for x in y = 10x + 150
y = 10 (52) + 150
y = 670 so 52 lessons will cost $670

Return to
problem

3. If John has $850, how many lesson can he buy?
Substitute 850 for y in y = 10x + 150
850 = 10x + 150
700 = 10x
x = 70 so $850 will pay for 70 lessons

Return to
problem

Solving Systems of Linear Equations
• Topics
• Solving systems graphically
• Solving systems by using tables
• Solving systems algebraically
• Substitution method
• Elimination method
• Systems with more than 2 variables
• Using matrix equations
• Using augmented matrices
• Applications of systems

Graphical Representation of
Linear Systems of Equations

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

Graphical Representation of
Systems of Linear Equations

y

y

y
y

Solving Systems Graphically
Click this link for a video of solving systems of linear equations
graphically.

Solving Systems Graphically
Click this link for a worksheet of solving systems of linear
equations graphically.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksh
eets/Systems%20of%20Equations%20Graphing.pdf

Solving Systems of Linear
Equations using tables
• Another method you can use to find the solution to a system
of linear equations is to make a table for each of the equations
and then compare to see if any of the ordered pairs of
solutions is in both tables of values.
• Example: y= 3x+2
and
y= x+6
x

3x+2

X

x+6

0

2

0

6

1

5

1

7

2

8

2

8

3

11

3

9

4

14

4

10

5

17

5

11

6

20

6

12

Solution

Solving Systems of Linear
Equations using tables
• Using tables of values to find a solution to a system of
equations is very inefficient. About the only time that you
would use this method is if the tables have already been
generated for you. Then you would just inspect the ordered
pairs to find the same values in each table, if they happen to
be listed. Example:
x

y

x

y

-3

-10

-3

5

-2

-8

-2

16/3

-1

-2

-1

14/3

0

-4

0

4

1

-2

1

10/3

2

0

2

8/3

3

2

3

2

The solution is:(3,2)

Solving Systems of Linear
Equations using tables
• However, with today’s technology. Your calculator can
generate tables of values very quickly.

From the graph above, you can tell that the x-coordinate of the intersection
is between -4 and -5. When you look at the table when x= -4 and -5, you’ll see the
Y-coordinates are 1 & 0 when x= -4 , and they are -1 & 0.25 when x is -5. This is close.

Solving Systems of Linear
Equations using tables
• Using your calculator you can refine the x-coordinates in you
table to begin to get closer to the actual point of intersection.

Notice in the table after several refinements to the x-coordinate, we have the
point of intersection rounded to 4 decimal places (-4.4444, 0.1111).

Solving Systems of Linear
Equations algebraically
•
•
•
•

Using Substitution
Using Elimination
Systems with more than 2 variables
Using Matrix Equations- If you have never worked with
matrices, then skip this section
• Using reduced row echelon form of augmented matrices- If
you have never worked with matrices, then skip this section.

Solving Systems of Linear
Equations algebraically
Substitution Method
• The concept of solving systems of linear equations using
substitution is this:
1. Take one of the equations and isolate one of the variables on one side of the
equation (Get a variable by itself on one side of the equals sign).
2. Now use the other equation and wherever the isolated variable is, substitute the
expression that equals that variable from the first equation for that variable.
3. Now you have an equation with only one variable. Solve this equation using
methods demonstrated in the “Solving Linear Equations” section of this
powerpoint.
4. Last, take the value solved-for from the previous step, substitute this value into
either equation and then solve for the other variable.
5. Your answer will typically be written as an ordered pair of numbers

Solving Systems of Linear
Equations algebraically
Substitution Method: Example
To solve a system of linear equations using substitution:
1.

Isolate one of the variables on one side of one of the two equations.



2.

2x + 5y = 7

x + 4y = 2
x = 2 - 4y

Go to the other equation and in place of the variable isolated in step 1, substitute the expression equal to that variable.

 2(2 - 4y) + 5y = 7
3.

Now simplify and solve the equation for the remaining variable.





4.

4 – 8y + 5y = 7
4 – 3y = 7
-3y = 3
Y = -1
Last take the solution from step 3 and substitute that value into either of the original 2 equations, then solve that
equation.

 X + 4y = 2

 X + 4(-1) = 2
 X–4=2
 X=6
5.

Your answer will be an ordered pair of numbers. Convention is to list them in alphabetical order.

 (x,y) = (6, -1)

Solving Systems of Linear
Equations algebraically
Click the link below to see a video using the substitution method

• Note: this video begins with a review of solving systems
graphically. If you drag the slider at the bottom of the video to
minute 1:37, that is when substitution begins.
http://www.noodle.org/learn/details/27646/s
olving-linear-systems-of-equations-usingsubstitution

Solving Systems of Linear
Equations algebraically
Practice problems using substitution

Note: I find this site to be better for just checking your answers
to a system of equations, rather than trying to follow their
solution. It is correct, but it may be difficult to follow all of their
steps.

http://www.webmath.com/solver2.html

Solving Systems of Linear
Equations algebraically
Elimination Method
1.
2.

Simplify each equation so that each equation is in standard form (ax + by = c).
Place the two equations so that all the like terms are in the same position and align
them vertically.
 Example:


3.

2x + 3y = 7
4x – 3y = 11

If you are lucky you will be able to either add or subtract the two equations with one of
the variables disappearing and what remains is an equation with only one variable.
 Example:
2x + 3y = 7

4x – 3y = 11
 Add vertically: 6x
= 18
 Solve for x:
x= 3

4.

Now substitute “3” for x into either of the two original equations and solve for y.





5.

2 (3) + 3y = 7 (the first equation from above)
6 + 3y = 7
3y = 1
Y=1/3

Solution: ( 3, 1/3)

Solving Systems of Linear
Equations algebraically
Elimination Method
1.

Often once the equations have been simplified into standard form and placed vertically aligning all the like
terms, you will not be able to just and or subtract the two equations to make one of the variables disappear.
Notice below, if you add or subtract the two equations you still have two variables. This is not good.
 Example:


add:

2.

12x + 3y = 9
4x – 5y = 9
16x - 2y = 18

subtract:

12x + 3y = 9
4x – 5y = 9
8x +8y = 0

When this happens, you need to multiply (or divide) every term of one or both of the equations by something
to force the coefficients on one of the variables to be opposites of each other (this way you will be able to add
the two equations to eliminate that variable).

 I have found that my students tend to make many less careless errors when adding than when subtracting. This
is why in step 2 above I suggest that the coefficients on one of the variables be opposites of each other.
3.

Notice that 12 (the coefficient on x in the top equation) is a multiple of 4 (the coefficient on x in the bottom
equation). So we want to multiply the second equation by “-3”.
 Example:




4.

12x + 3y = 9
12x + 3y = 9
-3(4x – 5y) = -3(9)
-12x + 15y = -27
Add like terms vertically:
18y = -18
Solve for x:
y = -1

Now substitute “-1” for y into either of the two original equations and solve for x.
 4x – 5(-1) = 9 (I chose the second equation from above)
 -4x + 5 = 9
 -4x = 4
 x= -1

5.

Solution: ( -1, -1)

Solving Systems of Linear
Equations algebraically
Elimination Method
 Here is an example where you need to do even more work to prepare the
equations before adding (or subtracting) them.
 2x + 5y = 14
 3x – 2y = -36
First decide which variable to eliminate, x or y.
 You could force the coefficients on x to be 6 and -6, or you could force the coefficients on y
to be 10 and -10. Your choice, either way works. I’m going to chose to force the coefficient
on y to be 10 in the top equation by multiplying both sides of the equation by 2, and then
force the coefficient on y in the second equation to be -10 by multiplying both sides of the
second equation by 5.
 2(2x) + 2(5y) = 2(14)
4x + 10y = 28
 5(3x) – 5(2y) =5(-36)
15x - 10y = -180

Now add:
19x
= -152

Solve for x:
x = -8
 Now solve for Y:
 2(-8) +5y =14
 -16 + 5y = 14

5y = 30

y= 6
 Solution: (-8, 6)

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below to watch a video.
 http://www.youtube.com/watch?v=s7S3oJ_0Uvo

Solving Systems of Linear
Equations algebraically
Elimination Method: Click the link below for practice problems
and solutions for solving linear systems using elimination.
 http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Systems%
20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
If you have an equation with three variables, this is still called a linear equation,
but the graph of it is not a line. Since there are three variables and each variable
represents a dimension, this means our graph will be a plane in 3-D.
If you have a system of three equations representing three planes, you could
have the intersection at one point such as the intersection of 2 corner walls and
the ceiling in a room. This is the ideal situation because we will get an ordered
triple (x,y,z) for the solution.
Other times we could get a line for the intersection of three planes. Think of the
pages of a book with the pages representing planes, and the intersection of
these planes is a line representing the binding of the book.
We could have three parallel planes that never intersect and have no points of
intersection. We are not going to consider all of the possibilities in this document.

Solving Systems of Linear
Equations algebraically
System of linear equations with 3 variables
*We have one-point of intersection(consistent &
independent) a point,
*We have an infinite set of points of intersection
(consistent and dependent) a line or a plane(not shown)
*We have no points of intersection (inconsistent).
*Below are three possible solutions for three planes.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables
 This is a good video showing how to solve a system of three equations with
three variables.
 http://www.youtube.com/watch?v=g6FQBfIwf3w

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically
Matrix Equations
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
This system is already in standard form.
Take the coefficients on X, Y, and Z and write them in a 3 by 3 matrix.
Notice that the coefficient on a
variable is understood to be one if no
other number is multiplied by the
variable.

Solving Systems of Linear
Equations algebraically
Elimination Method: system with 3 variables

Solving Systems of Linear
Equations algebraically

Matrix Equations: This section assumes that you have had
some experience working with matrices. If you have not, then
skip this part.
Write this system using matrix equations:
2x + y + 3z = 2
x + y + 8z = 2
x+y+z=3
First the equations need to be in standard form:
AX + BY = C (If there are two variables X & Y.)

AX + BY + CZ = D (if there are three variables, X, Y and Z.)
AX + BY + CZ +DW = E (If there are 4 variables, X, Y, Z, W.)
and so on.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Next multiply this matrix by the 3 by 1
variable matrix and set this matrix
equation equal to the 3 by 1 matrix
created from the constants in the original
equations.

Solving Systems of Linear
Equations algebraically
Matrix Equations

Note: How to find inverses of matrices is not included in this tutorial. However, if
you have a graphics calculator, it can easily find the inverse for you. If it exists.

Solving Systems of Linear
Equations algebraically
Matrix Equations

0

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations

Solving Systems of Linear
Equations algebraically
Matrix Equations: Practice problems

This is the same link from earlier. The only difference is that you now may
want to solve these problems using matrix equations.
http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/
Systems%20of%20Equations%20Elimination.pdf

Solving Systems of Linear
Equations algebraically

Augmented Matrices: You may want to skip this part of the unit.
This may be beyond the scope of this unit. This is a quick way to
solve systems, but in no way does it cover all of the concepts
needed to understand and use matrices.

Note: The second equation has no “Y”,
so we enter a “0” for the coefficient in the matrix.

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Reduced Row Echelon Form(rref) of Augmented Matrices:
Click below to watch a video using augmented matrices to
solve a 3 by 3 system of linear equations. This video is worth
watching even if you don’t want to know how to solve
systems by hand using augmented matrices. The person
in the video clearly loves what he is doing.
http://youtu.be/oCygbOvQqtw

Solving Systems of Linear
Equations algebraically
Augmented Matrices using your graphics calculator

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Solving Systems of Linear
Equations algebraically
Augmented Matrices

Applications of Systems of
Linear Equations
Why study systems of linear equations?
• Systems of linear equations are everywhere. When studying
mathematics, economics, sciences, business, etc. you will
encounter situations where needing to answer a problem will
involve solving a system of equations.
• Examples: Rate of work problems, Percent mixture problems,
Rate-time-distance problems, Problems finding when the cost
of multiple scenarios is the least, money and age problems,
etc.
• Solving systems is a nice skill to have, but it even more
valuable when you have the ability to use that skill to set-up
and solve real world problems.

Applications of Systems of
Linear Equations
Application of Linear Sytems
Suppose the local market place sells a fruit smoothie for $2.50 and a cup
of hot chocolate for $2.00. On Thursday, Malinda sold 30 more fruit
smoothies than hot chocolates for a total of $178.50 worth of fruit
smoothies and hot chocolate. How many cups of each did Malinda sell?
First, we need to underline the important information.
Suppose the local market place sells a fruit smoothie for $2.50
and a cup of hot chocolate for $2.00. On Thursday, Malinda sold
30 more fruit smoothies than hot chocolates for a total of
$178.50 worth of fruit smoothies and hot chocolate. How many
cups of each did Malinda sell?

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
Second, we must define our variables.
Let “c” represent the number of cups of hot chocolate and “s” represent the
number of fruit smoothies.
Next, write equations to represent the underlined information from step 1.
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothie for $2.50
s = 30 + c
(30 more than hot chocolates)
c + s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
c
+
s = total number of cups sold
# of cups of
hot chocolate

# of smoothies

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
c = # of cups of hot chocolate for $2.00
s = # of fruit smoothies for $2.50
s = 30 + c
c + s = total number of cups sold
$2c + $2.50s = $178.50 (for a total of $178.50 worth of fruit smoothies and hot chocolate)
2 c + 2.50(30 + c ) = 178.50
2c + 2.50(30) + 2.50 c = 178.50
2c + 2.50c + 75 = $178.50
4.50c + 75= 178.50
4.50c + 75 -75= 178.50 -75
4.50c = 103.50
c = 23

Solve for c
Use the distributive property
Simplify
Combine like terms
Subtract 75
Divide by 4.50
Number of cups of hot chocolate

Applications of Systems of
Linear Equations
Application of Linear Systems- continued
How many cups of each did Malinda sell?
c = 23 cups of hot chocolate
s = 30 + c
(30 more smoothies than hot chocolates)
s = 30 + 23
s = 53 fruit smoothies

Malinda sold 23 cups of hot chocolate and 53 fruit smoothies
Check your answer: $2 (23) + $2.50 (53) = $178.50
$46 + $132.5 = $178.50

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Rate-Time Distance Application

Applications of Systems of
Linear Equations
Click the links below for practice problems
• This site gives 6 problems to solve followed by a second page of the
solutions. It is good for practice.
http://www.ohiorc.org/orc_documents/orc/for_mathematics/tutorials/
35_selfcheck.pdf

• This site shows 3 examples of applications.
1. The first example involves money and tickets,
2. the second example beginning at minute 4:00 is a percent mixture
problem,
3. and the third example beginning at minute 8:00 is rate-time-distance
problem.
http://www.youtube.com/watch?v=FRaJv2Faass